The equation of a circle $C$ is $x^2+y^2-4y+3 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2) + (y^2-4y) = -3$ $(x^2) + (y^2-4y+4) = -3 + 0 + 4$ $x^2 + (y-2)^{2} = 1 = 1^2$ Thus, $(h, k) = (0, 2)$ and $r = 1$.